y′ − (y/(x^2 −1)) = y^2


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Question Number 103776 by bemath last updated on 17/Jul/20

$y^{'} - \frac{y}{x^{2} - 1} = y^{2}$
	Answered by bemath last updated on 17/Jul/20
	$set v = y^(−1) ⇒(dv/dx) = −y^(−2) (dy/dx) (dy/dx) = −y^2 (dv/dx) then −y^2 (dv/dx)−(y/(x^2 −1)) = y^2 (dv/dx) + (v/(x^2 −1)) = −1 integrating factor j(x) = e^(∫ (dx/((x−1)(x+1)))) = e^((1/2)∫(dx/(x−1))−(dx/(x+1))) j(x)= e^((1/2)ln (((x−1)/(x+1)))) = (√((x−1)/(x+1))) we get v(x) = ((∫ (√((x−1)/(x+1))).(−1) dx+C)/(√((x−1)/(x+1)))) v(x) =−(√((x+1)/(x−1))) {∫((√(x−1))/(√(x+1))) dx + C} (1/y) = −(√((x+1)/(x−1))) {∫ ((√(x−1))/(√(x+1))) dx + C } (1/y) = −(√((x+1)/(x−1))) {(√(x^2 −1))−2tan^(−1) ((√((x+1)/(x−1))))+C} (1/y)=−x−1+2(√((x+1)/(x−1)))tan^(−1) ((√((x+1)/(x−1))))−C(√((x+1)/(x−1)))$
	$s e t v = y^{- 1} \Rightarrow \frac{d v}{d x} = - y^{- 2} \frac{d y}{d x}$ $\frac{d y}{d x} = - y^{2} \frac{d v}{d x}$ $t h e n - y^{2} \frac{d v}{d x} - \frac{y}{x^{2} - 1} = y^{2}$ $\frac{d v}{d x} + \frac{v}{x^{2} - 1} = - 1$ $i n t e g r a t i n g f a c t o r$ $j (x) = e^{\int \frac{d x}{(x - 1) (x + 1)}} = e^{\frac{1}{2} \int \frac{d x}{x - 1} - \frac{d x}{x + 1}}$ $j (x) = e^{\frac{1}{2} \ln (\frac{x - 1}{x + 1})} = \sqrt{\frac{x - 1}{x + 1}}$ $w e g e t v (x) = \frac{\int \sqrt{\frac{x - 1}{x + 1}} . (- 1) d x + C}{\sqrt{\frac{x - 1}{x + 1}}}$ $v (x) = - \sqrt{\frac{x + 1}{x - 1}} {\int \frac{\sqrt{x - 1}}{\sqrt{x + 1}} d x + C}$ $\frac{1}{y} = - \sqrt{\frac{x + 1}{x - 1}} {\int \frac{\sqrt{x - 1}}{\sqrt{x + 1}} d x + C}$ $\frac{1}{y} = - \sqrt{\frac{x + 1}{x - 1}} {\sqrt{x^{2} - 1} - {2 \tan}^{- 1} (\sqrt{\frac{x + 1}{x - 1}}) + C}$ $\frac{1}{y} = - x - 1 + 2 \sqrt{\frac{x + 1}{x - 1}} \tan^{- 1} (\sqrt{\frac{x + 1}{x - 1}}) - C \sqrt{\frac{x + 1}{x - 1}}$
	Commented by Ar Brandon last updated on 17/Jul/20
	What's the secret to knowing the right change of variable to use ? ��
	Commented by bobhans last updated on 17/Jul/20

	$(⌢∵⌢) \int \frac{\sqrt{x - 1}}{\sqrt{x + 1}} d x = I$ $I = \sqrt{(x - 1) (x + 1)} - {2 \tan}^{- 1} \sqrt{\frac{x + 1}{x - 1}} + k$
	Commented by bemath last updated on 17/Jul/20

	$B e r n o u l l i e q s i r$
	Commented by Coronavirus last updated on 17/Jul/20

	${i t}^{'} s a B e r n o u l l i e q u a t i o n$
	Answered by mathmax by abdo last updated on 17/Jul/20
	$e⇒y^′ −(1/(x^2 −1))y =y^2 ⇒(y^′ /y^2 )−(1/(x^2 −1))×(1/y) =1 let z =(1/y) ⇒y =(1/z) ⇒y^′ =−(z^′ /z^2 ) e⇒−(z^′ /z^2 )×z^2 −(1/(x^2 −1))×z =1 ⇒−z^′ −(1/(x^2 −1))z =1 ⇒z^′ +(z/(x^2 −1)) =−1 e^, he→z^′ =−(z/(x^2 −1)) ⇒(z^′ /z) =−(1/(x^2 −1)) ⇒ln∣z∣ =−∫ (dx/(x^2 −1)) =−(1/2)∫((1/(x−1))−(1/(x+1)))dx =−(1/2)ln∣((x−1)/(x+1))∣ +c =(1/2)ln∣((x+1)/(x−1))∣ +c ⇒z =k (√(∣((x+1)/(x−1))∣)) solution on w ={x/((x+1)/(x−1))>0} mvc method →z^′ =k^′ (√((x+1)/(x−1))) +k ×(((((x+1)/(x−1)))^′ )/(2(√((x+1)/(x−1))))) =k^′ (√((x+1)/(x−1)))+(k/2)×((−2)/((x−1)^2 (√((x+1)/(x−1))))) e^′ ⇒ k^′ (√((x+1)/(x−1)))−(k/((x−1)^2 (√((x+1)/(x−1))))) +(1/(x^2 −1))×k(√(((x+1)/(x−1)) ))=−1 ⇒ k^′ =−(√((x−1)/(x+1))) ⇒k =−∫ (√((x−1)/(x+1)))dx changement (√((x−1)/(x+1)))=t give ((x−1)/(x+1))=t^2 ⇒x−1 =t^2 x+t^2 ⇒(1−t^2 )x =t^2 +1 ⇒x =((t^2 +1)/(1−t^2 )) dx =((2t(1−t^2 )−(t^2 +1)(−2t))/((1−t^2 )^2 ))dt =((2t−2t^3 +2t^3 +2t)/((t^2 −1)^2 ))dt =((4tdt)/((t^2 −1)^2 )) ⇒ ∫ (√((x−1)/(x+1)))dx =4 ∫ ((tdt)/((t^2 −1)^2 )) =4 ∫ ((tdt)/((t−1)^2 (t+1)^2 )) =4 ∫ ((tdt)/((((t−1)/(t+1)))^2 (t+1)^4 )) we do the changement ((t−1)/(t+1))=z ⇒t−1 =zt+z ⇒ (1−z)t=1+z ⇒t =((1+z)/(1−z)) ⇒(dt/dz) =((1−z+(1+z))/((1−z)^2 )) =(2/((1−z)^2 )) t+1 =((1+z)/(1−z)) +1 =((1+z+1−z)/(1−z)) =(2/(1−z)) ⇒I= ∫ ((1+z)/(1−z))×(2/((1−z)^2 z^2 ((2/(1−z)))^4 ))dz =(1/8)∫ (((1+z)(1−z)^4 )/((1−z)^3 ))dz =(1/8)∫ (1−z^2 )dz =(1/8)(z−(z^3 /3)) +c =(1/8)(((t−1)/(t+1)))−(1/(24))(((t−1)/(t+1)))^3 +c with t =(√((x−1)/(x+1))) y(x)=(1/(z(x)))$
	$e \Rightarrow y^{^{'}} - \frac{1}{x^{2} - 1} y = y^{2} \Rightarrow \frac{y^{^{'}}}{y^{2}} - \frac{1}{x^{2} - 1} \times \frac{1}{y} = 1 let z = \frac{1}{y} \Rightarrow y = \frac{1}{z}$ $\Rightarrow y^{^{'}} = - \frac{z^{^{'}}}{z^{2}}$ $e \Rightarrow - \frac{z^{^{'}}}{z^{2}} \times z^{2} - \frac{1}{x^{2} - 1} \times z = 1 \Rightarrow - z^{^{'}} - \frac{1}{x^{2} - 1} z = 1 \Rightarrow z^{^{'}} + \frac{z}{x^{2} - 1} = - 1 e^{,}$ $he \to z^{^{'}} = - \frac{z}{x^{2} - 1} \Rightarrow \frac{z^{^{'}}}{z} = - \frac{1}{x^{2} - 1} \Rightarrow \ln ∣ z ∣ = - \int \frac{dx}{x^{2} - 1} = - \frac{1}{2} \int (\frac{1}{x - 1} - \frac{1}{x + 1}) dx$ $= - \frac{1}{2} \ln ∣ \frac{x - 1}{x + 1} ∣ + c = \frac{1}{2} \ln ∣ \frac{x + 1}{x - 1} ∣ + c \Rightarrow z = k \sqrt{∣ \frac{x + 1}{x - 1} ∣}$ $solution on w = {x / \frac{x + 1}{x - 1} > 0} mvc method \to z^{^{'}} = k^{^{'}} \sqrt{\frac{x + 1}{x - 1}}$ $+ k \times \frac{{(\frac{x + 1}{x - 1})}^{^{'}}}{2 \sqrt{\frac{x + 1}{x - 1}}} = k^{^{'}} \sqrt{\frac{x + 1}{x - 1}} + \frac{k}{2} \times \frac{- 2}{{(x - 1)}^{2} \sqrt{\frac{x + 1}{x - 1}}}$ $e^{^{'}} \Rightarrow k^{^{'}} \sqrt{\frac{x + 1}{x - 1}} - \frac{k}{{(x - 1)}^{2} \sqrt{\frac{x + 1}{x - 1}}} + \frac{1}{x^{2} - 1} \times k \sqrt{\frac{x + 1}{x - 1}} = - 1 \Rightarrow$ $k^{^{'}} = - \sqrt{\frac{x - 1}{x + 1}} \Rightarrow k = - \int \sqrt{\frac{x - 1}{x + 1}} dx changement \sqrt{\frac{x - 1}{x + 1}} = t give$ $\frac{x - 1}{x + 1} = t^{2} \Rightarrow x - 1 = t^{2} x + t^{2} \Rightarrow (1 - t^{2}) x = t^{2} + 1 \Rightarrow x = \frac{t^{2} + 1}{1 - t^{2}}$ $dx = \frac{2 t (1 - t^{2}) - (t^{2} + 1) (- 2 t)}{{(1 - t^{2})}^{2}} dt = \frac{2 t - {2 t}^{3} + {2 t}^{3} + 2 t}{{(t^{2} - 1)}^{2}} dt = \frac{4 tdt}{{(t^{2} - 1)}^{2}} \Rightarrow$ $\int \sqrt{\frac{x - 1}{x + 1}} dx = 4 \int \frac{tdt}{{(t^{2} - 1)}^{2}} = 4 \int \frac{tdt}{{(t - 1)}^{2} {(t + 1)}^{2}}$ $= 4 \int \frac{tdt}{{(\frac{t - 1}{t + 1})}^{2} {(t + 1)}^{4}} we do the changement \frac{t - 1}{t + 1} = z \Rightarrow t - 1 = zt + z \Rightarrow$ $(1 - z) t = 1 + z \Rightarrow t = \frac{1 + z}{1 - z} \Rightarrow \frac{dt}{dz} = \frac{1 - z + (1 + z)}{{(1 - z)}^{2}} = \frac{2}{{(1 - z)}^{2}}$ $t + 1 = \frac{1 + z}{1 - z} + 1 = \frac{1 + z + 1 - z}{1 - z} = \frac{2}{1 - z} \Rightarrow I = \int \frac{1 + z}{1 - z} \times \frac{2}{{(1 - z)}^{2} z^{2} {(\frac{2}{1 - z})}^{4}} dz$ $= \frac{1}{8} \int \frac{(1 + z) {(1 - z)}^{4}}{{(1 - z)}^{3}} dz = \frac{1}{8} \int (1 - z^{2}) dz$ $= \frac{1}{8} (z - \frac{z^{3}}{3}) + c = \frac{1}{8} (\frac{t - 1}{t + 1}) - \frac{1}{24} {(\frac{t - 1}{t + 1})}^{3} + c with t = \sqrt{\frac{x - 1}{x + 1}}$ $y (x) = \frac{1}{z (x)}$
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